Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 43

\[\boxed{\mathbf{43.}}\]

\[\textbf{а)}\ \frac{\sqrt{x^{2} + 1} - \left| \cos x \right|}{\sqrt{2x} + \left| \cos x \right|} >\]

\[> \frac{\sqrt{2x} - |\cos x|}{\sqrt{x^{2} + 1} + |\cos x|}\]

\[x \geq 0;\]

\[M = \lbrack 0; + \infty).\]

\[x^{2} + 1 - cos^{2}x > 2x - cos^{2}x\]

\[x^{2} - 2x + 1 > 0\]

\[(x - 1)^{2} > 0\]

\[x \neq 1.\]

\[Решение\ неравенства:\]

\[x \in \lbrack 0;1) \cup (1; + \infty).\]

\[Ответ:x \in \lbrack 0;1) \cup (1; + \infty).\]

\[\textbf{б)}\ \frac{\sqrt{x^{2} - 4} - \left| \sin x \right|}{\sqrt{3x} + \left| \sin x \right|} <\]

\[< \frac{\sqrt{3x} - \left| \sin x \right|}{\sqrt{x^{2} + 4} + \left| \sin x \right|}\]

\[x \geq 0;\]

\[x^{2} - 4 \geq 0\]

\[x^{2} \geq 4\]

\[x \leq - 2;\ \ x \geq 2.\]

\[M = \lbrack 2; + \infty).\]

\[x^{2} - 4 - sin^{2}x < 3x - sin^{2}x\]

\[x^{2} - 3x - 4 < 0\]

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\ \ x_{2} = - 1;\]

\[(x + 1)(x - 4) < 0\]

\[- 1 < x < 4.\]

\[Решение\ неравенства:\]

\[x \in \lbrack 2;4).\]

\[Ответ:x \in \lbrack 2;4).\]